Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
class Solution {
 public:
  Solution(vector<pair<long long, long long>>& fences, long long a, long long b,
           long long sy)
      : segs(fences), pl(a), pr(b), py(sy) {
    int n = segs.size();
    sums.resize(n + 1);
    for (int i = 0; i < n; ++i) {
      sums[i + 1] = sums[i] + segs[i].second - segs[i].first;
    }
  }
  long double query(long long x, long long y) {
    long double fl = x + ((long double)y * (pl - x) / (y - py));
    long double fr = x + ((long double)y * (pr - x) / (y - py));
    int l = getGE(fl);
    int r = getGE(fr) - 1;
    long double res = 0;
    if (r >= l) {
      res = sums[r] - sums[l];
    }
    if (l - 1 >= 0 && r != l - 1) {
      res += max(0.0L, (long double)segs[l - 1].second - fl);
      if (segs[l - 1].first > fl) {
        cout << l - 1 << " " << segs[l - 1].first << " " << fl << endl;
      }
    }
    if (r >= 0) {
      res += max(0.0L, min(fr, (long double)segs[r].second) -
                           max(fl, (long double)segs[r].first));
    }
    res *= (long double)(y - py) / y;
    return res;
  }

 private:
  long long pl;
  long long pr;
  long long py;
  vector<pair<long long, long long>> segs;
  vector<long long> sums;
  int getGE(long double x) {
    auto it =
        lower_bound(segs.begin(), segs.end(), make_pair((long long)x, -1LL));
    int i = it - segs.begin();
    i = max(i - 2, 0);
    while (i < segs.size()) {
      if (segs[i].first >= x) {
        return i;
      }
      ++i;
    }
    return i;
  }
};
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int sy, a, b;
  cin >> sy >> a >> b;
  int n;
  cin >> n;
  vector<pair<long long, long long>> segs;
  segs.reserve(n);
  for (int i = 0, l = 0, r = 0; i < n; ++i) {
    cin >> l >> r;
    segs.emplace_back(l, r);
  }
  Solution sol(segs, a, b, sy);
  int q;
  cin >> q;
  cout << fixed << setprecision(15);
  for (int i = 0, x = 0, y = 0; i < q; ++i) {
    cin >> x >> y;
    cout << sol.query(x, y) << endl;
  }
  return 0;
}